4.1 Related Rates

Big idea

Until now a derivative answered a question about one function of one variable. Here two or more quantities change at once, and they are not independent: a geometric or physical relation ties them together at every instant. If you know how fast one of them is moving, the relation forces a value on the speed of the other.

The mechanism is implicit differentiation with time as the variable everything depends on. Suppose a relation like $x^2 + y^2 = 100$ holds for all $t$, with $x$ and $y$ both functions of time. Then the two sides are the same function of $t$, so their derivatives agree. Differentiating the left side needs the chain rule on each term, because $x$ is not the variable you are differentiating with respect to; it is a function of the variable. That is where each $dx/dt$ factor comes from.

The result is a second equation, this one relating rates instead of positions. It holds at every instant, so you may substitute the particular values of the moment you care about into it. Substituting those values before you differentiate is the error that ruins the problem: a quantity that you pin to a number stops being a function of time, and its derivative comes out zero.

So the work splits cleanly. First model the geometry and get a relation that is true for all time. Then differentiate with respect to time. Only then read off the instant you were asked about.

Decoder

If two changing quantities satisfy the same equation at every instant, then differentiating that equation with respect to time produces an equation their rates must satisfy at every instant.

The phrase carrying the content is “at every instant.” A relation that happens to be true only at the single moment in question tells you nothing about rates. The ladder’s length stays $10$ for all time, so it is a constant and differentiates to zero. The distance from the wall is $6$ only right now, so it must stay a letter until after the differentiation.

Every term is differentiated by the chain rule, whether or not you write the intermediate step. The derivative of $y^3$ with respect to $t$ is $3y^2 \frac{dy}{dt}$, because $y^3$ is the cube function composed with the function $y(t)$. Terms with both variables need the product rule as well: the derivative of $xy$ is $x\frac{dy}{dt} + y\frac{dx}{dt}$.

The classic mistake, then, is substituting early. The second classic mistake is a sign: a shrinking quantity has a negative rate, and if you write the rate as a positive number the answer comes out with the wrong direction.

Definitions and results

Rate of change with respect to time. If a quantity $Q$ varies with time, $\frac{dQ}{dt}$ is its instantaneous rate of change, measured in units of $Q$ per unit time. Positive means growing, negative means shrinking.

The governing relation. A related rates problem supplies an equation among the changing quantities that holds at all times. Common sources are the Pythagorean theorem, area and volume formulas, similar triangles, and the trigonometric ratios in a right triangle.

Differentiating the relation. Treat each variable as a function of $t$ and differentiate both sides with respect to $t$. Every appearance of a variable contributes a chain rule factor:

$$ \frac{d}{dt}\big[f(x)\big] = f'(x)\,\frac{dx}{dt} $$

Constants versus instantaneous values. A number that never changes may be substituted at any time. A number that describes the present moment may be substituted only after differentiating.

Eliminating a variable with a constraint. When a fixed proportion links two quantities, use it before differentiating to cut the number of variables. In a cone whose radius is always half its height, write $r = h/2$ and rewrite the volume in terms of $h$ alone.

A standard shape. For a right triangle with legs $x$ and $y$ and hypotenuse $z$, differentiating $x^2 + y^2 = z^2$ gives

$$ x\frac{dx}{dt} + y\frac{dy}{dt} = z\frac{dz}{dt} $$

after cancelling the factor of $2$. This one line covers sliding ladders, separating vehicles, and shadow problems.

Units as a check. The units of the answer are forced by the relation. If $V$ is in cubic feet and $t$ in minutes, then $dV/dt$ is cubic feet per minute and $dh/dt$ comes out in feet per minute. A mismatch means an algebra slip.

Worked examples

A ladder sliding down a wall

A $10$ ft ladder leans against a vertical wall. Its base is pulled away from the wall at $2$ ft/s. How fast is the top sliding down when the base is $6$ ft from the wall?

Let $x$ be the distance from wall to base and $y$ the height of the top. The ladder length is constant, so $x^2 + y^2 = 100$ for all $t$. Differentiate:

$$ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \qquad \Longrightarrow \qquad \frac{dy}{dt} = -\frac{x}{y}\cdot\frac{dx}{dt} $$

At the instant asked, $x = 6$, so $y = \sqrt{100 - 36} = 8$, and $dx/dt = 2$. Then

$$ \frac{dy}{dt} = -\frac{6}{8}(2) = -\frac{3}{2} $$

The top is falling at $1.5$ ft/s. The sign is negative because the height is decreasing, which is what a downward slide should produce.

A balloon being inflated

Air enters a spherical balloon at $100$ cm$^3$/s. How fast is the radius growing when the radius is $5$ cm?

Volume and radius are tied by $V = \frac{4}{3}\pi r^3$ at all times. Differentiating with respect to $t$,

$$ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} $$

Put in $dV/dt = 100$ and $r = 5$:

$$ 100 = 4\pi(25)\frac{dr}{dt} \qquad \Longrightarrow \qquad \frac{dr}{dt} = \frac{1}{\pi} $$

The radius grows at $1/\pi$ cm/s, about $0.32$ cm/s. Notice the structure: the factor $4\pi r^2$ is the surface area, so a fixed inflow spreads over a larger skin as the balloon grows and the radius responds more slowly.

Draining a conical tank

A tank is an inverted cone $12$ ft deep whose top rim has radius $6$ ft. Water drains out at $2$ ft$^3$/min. How fast is the depth dropping when the water is $4$ ft deep?

The water forms a similar cone, so its radius and depth satisfy $r/h = 6/12$, that is $r = h/2$. Substitute before differentiating so only one variable remains:

$$ V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{2}\right)^{2} h = \frac{\pi h^3}{12} $$

Now differentiate with respect to $t$:

$$ \frac{dV}{dt} = \frac{\pi h^2}{4}\cdot\frac{dh}{dt} $$

With $dV/dt = -2$ (draining, so negative) and $h = 4$:

$$ -2 = \frac{\pi(16)}{4}\cdot\frac{dh}{dt} = 4\pi\frac{dh}{dt} \qquad \Longrightarrow \qquad \frac{dh}{dt} = -\frac{1}{2\pi} $$

The water level drops at $1/(2\pi)$ ft/min, roughly $0.16$ ft/min.

Two vehicles separating

A car leaves an intersection heading east at $30$ mph. A second car leaves the same intersection heading north at $40$ mph. How fast is the gap between them growing when the first is $3$ mi east and the second is $4$ mi north?

With $x$ east, $y$ north and $z$ the gap, $x^2 + y^2 = z^2$ holds at all times. At the instant in question $z = \sqrt{9 + 16} = 5$. Using the differentiated form,

$$ \frac{dz}{dt} = \frac{x\frac{dx}{dt} + y\frac{dy}{dt}}{z} = \frac{3(30) + 4(40)}{5} = \frac{250}{5} = 50 $$

The gap grows at $50$ mph. Sanity check the size: the separation rate should sit between the two speeds and below their sum, and $30 < 50 < 70$. It equals $50$ exactly here because the position vector happens to point the same way as the velocity vector, both in the ratio $3:4$.

Practice

Start with the full modelling loop. Each drill gives you a geometric or physical relation, one known rate, and one instant, and asks for the unknown rate.

Then drill the differentiation step on its own. These ask you to differentiate a relation implicitly, which is the same operation with $t$ replaced by another variable.

Quiz

Five items on setting up the relation, differentiating it with respect to time, and reading off the rate at a given instant.